#### Question

If n is positive integer, show that n(n^2-1)(n^2-5n+26) is divisible by 120

Question

If n is positive integer, show that n(n^2-1)(n^2-5n+26) is divisible by 120

n(n^2-1)(n^2-5n+26)

=(n-1)n(n+1)[(n-2)(n-3)+20]

=(n-2)(n-2)(n-1)n(n+1)+20(n-1)n(n+1)

The first term is 5 consecutive integers, which must have 2 even numbers and one of them must be multiples of 4. There also must be a number that is divisible by 3. There also must be a number that is divisible by 5.

Therefore it must be divisible by 2\times 4\times 3\times 5=120

The second term is the product of 20 and 3 consecutive numbers, in which there must be 1 even number and a multiple of 3.

Therefore it must be divisible byy 20\times2\times3=120